# Graph, girth and expanders

In the book “Elementary number theory, group theory and Ramanujan graphs“, Sarnak et. al. gave an elementary construction of expander graphs. We decided to go through the construction in the small seminar and I am recently assigned to give a talk about the girth estimate of such graphs.

Given graph (finite and undirected) $G$, we will denote the set of vertices by $V(G)$ and the set of edges $E(G) \subseteq V(G)^2$. The graph is assumed to be equipped with the standard metric where each edge has length $1$.

The Cheeger constant (or isoperimetric constant of a graph, see this pervious post) is defined to be:

$\displaystyle h(G) = \inf_{S\subseteq V_G} \frac{|\partial S|}{\min\{|S|, |S^c|\}}$

Here the notation $\partial S$ denote the set of edges connecting an element in $S$ to an element outside of $S$.

Note that this is indeed like our usual isoperimetric inequalities since it’s the smallest possible ratio between size of the boundary and size of the set it encloses. In other words, this calculates the most efficient way of using small boundary to enclose areas as large as possible.

It’s of interest to find graphs with large Cheeger constant (since small Cheeger constant is easy to make: take two large graphs and connect them with a single edge).

It’s also intuitive that as the number of edges going out from each vertice become large, the Cheeger constant will become large. Hence it make sense to restrict ourselves to graphs where there are exactly $k$ edges shearing each vertex, those are called $k$-regular graphs.

If you play around a little bit, you will find that it’s not easy to build large k-regular graphs with Cheeger constant larger than a fixed number, say, $1/10$.

Definition: A sequence of k-regular graphs $(G_i)$ where $|V_{G_i}| \rightarrow \infty$ is said to be an expander family if there exists constant $c>0$ where $h(G_i) \geq c$ for all $i$.

By random methods due to Erdos, we can prove that expander families exist. However an explicit construction is much harder.

Definition: The girth of $G$ is the smallest non-trivial cycle contained in $G$. (Doesn’t this smell like systole? :-P)

In the case of trees, since it does not contain any non-trivial cycle, define the girth to be infinity.

The book constructs for us, given pair $p, q$ of primes where $p$ is large (but fixed) and $q \geq p^8$, a graph $(p+1)$-regular graph$X^{p,q}$ with

$\displaystyle h(X^{p,q}) \geq \frac{1}{2}(p+1 - p^{5/6 + \varepsilon} - p^{1/6-\varepsilon})$

where $0 < \varepsilon < 1/6$.

Note that the bound is strictly positive and independent of $q$. Giving us for each $p$, $(X^{p,q})$ as q runs through primes larger than $p^8$ is a $(p+1)$-regular expander family.

In fact, this constructs for us an infinite family of expander families: a $(k+1)$-regular one for each prime $k$ and the uniform bound on Cheeger constant gets larger as $k$ becomes larger.

One of the crucial step in proving this is to bound the girth of the graph $X^{p,q}$, i.e. they showed that $g(X^{p,q}) \geq 2 \log_p(q)$ and if the quadratic reciprocity $(\frac{p}{q}) = -1$ then $g(X^{p,q}) \geq 4 \log_p(q) - \log_p(4)$. This is what I am going to do in this post.

Let $\mathbb H ( \mathbb Z)$ be the set of quaternions with $\mathbb Z$ coefficient, i.e.

$\mathbb H ( \mathbb Z) = \{ a+bi+cj+dk \ | \ a,b,c,d \in \mathbb Z \}$

Fix odd prime $p$, let

$\Lambda' = \{ \alpha \in \mathbb H(\mathbb Z) \ | \ \alpha \equiv 1 (mod 2) \}$

$\cup \ \{\alpha \in \mathbb H(\mathbb Z) \ | \ N(\alpha) = p^n \ \mbox{and} \ \alpha \equiv i+j+k (mod 2) \}$

where the norm $N$ on $\mathbb H(\mathbb Z)$ is the usual $N(a+bi+cj+dk) = a^2 + b^2 +c^2+d^2$.

$\Lambda'$ consists of points with only odd first coordinate or points lying on spheres of radius $\sqrt{p^n}$ and having only even first coordinate. One can easily check $\Lambda'$ is closed under multiplication.

Define equivalence relation $\sim$ on $\Lambda'$ by

$\alpha \sim \beta$ if there exists $m, n \in \mathbb{N}$ s.t. $p^m \alpha = \pm p^n \beta$.

Let $\Lambda = \Lambda' / \sim$, let $Q: \Lambda' \rightarrow \Lambda$ be the quotient map.

Since we know $\alpha_1 \sim \alpha_2, \beta_1 \sim \beta_2 \Rightarrow \alpha_1\beta_1 \sim \alpha_2\beta_2$, $\Lambda$ carries an induced multiplication with unit.

In elementary number theory, we know that the equation $a^2+b^2+c^2+d^2 = p$ has exactly $8(p+1)$ integer solutions. Hence the sphere of radius $p$ in $\mathbb H(\mathbb Z)$ contain $8(p+1)$ points.

In each $4$-tuple $(a,b,c,d)$ exactly one is of a different parity from the rest, depending on whether $p\equiv1$ or $3 (mod 4)$. Restricting to solutions where the first coordinate is non-negative, having different parity from the rest (in case the first coordinate is $0$, we pick one of the two solutions $\alpha, -\alpha$ to be canonical), this way we obtain exactly $p+1$ solutions.

Let $S'_p = \{ \alpha_1, \bar{\alpha_1}, \cdots, \alpha_k, \bar{\alpha_k}, \beta_1, \cdots, \beta_l \}$ be this set of $p+1$ points on the sphere. Note that the $\beta$s represent the solutions where the first coordinate is exactly $0$.

Check that $S'_p$ generates $\Lambda'$.

We have $\alpha_i \bar{\alpha_i} = p$ and $-\beta_j^2 = p$. By definition $S'_p \subseteq \Lambda'$ and $Q$ is injective on $S'_p$. Let $S_p = Q(S'_p)$.

Consider the Cayley graph $\mathcal G (\Lambda, S_p)$, this is a $(p+1)$-regular graph. Since $S_p$ generares $\Lambda$, $\mathcal G (\Lambda, S_p)$ is connected.

Claim: $\mathcal G (\Lambda, S_p)$ is a tree.

Suppose not, let $(v_0, v_1, \cdots, v_k=v_0)$ a non-trivial cycle. $k \geq 2$. Since $\mathcal G$ is a Cayley graph, we may assume $v_0 = e$.

Hence $v_1 = \gamma_1, \ v_2 = \gamma_1\gamma_2, \cdots, v_k = \gamma_1 \cdots \gamma_k$, for some $\gamma_1, \cdots, \gamma_k \in S_p$.

Since $v_{i-1} \neq v_{i+1}$ for all $1\leq i \leq k-1$, the word $\gamma_1, \cdots, \gamma_k$ cannot contain either $\alpha_i\bar{\alpha_i}$ or $\beta_i^2$, hence cannot be further reduced.

$\gamma_1, \cdots, \gamma_k = e$ in $\Lambda$ means for some $m, n$ we have

$\pm p^n \gamma_1, \cdots, \gamma_k = p^m$.

Since every word in $\Lambda'$ with norm $N(\alpha) = p^k$ must have a unique factorization $\alpha = \pm p^r w_m$ where $w_m$ is a reduces word of length $m$ in $S'_p$ and $2r+m = k$.

Contradiction. Establishes the claim.

Now we reduce the group $\mathbb H (\mathbb Z)$ mod $q$:

$\pi_q: \mathbb H (\mathbb Z) \rightarrow \mathbb H (\mathbb{F}_q)$

One can check that $\pi_q(\Lambda') = \mathbb H (\mathbb{F}_q)^\times$.

Let $Z_q = \{ \alpha \in \mathbb H (\mathbb{F}_q)^\times \ | \ \alpha = \bar{\alpha} \}$, $Z_q < \mathbb H (\mathbb{F}_q)^\times$ is a central subgroup.

For $\alpha, beta \in \Lambda'$, $\alpha \sim \beta \Rightarrow \pi_q(\alpha)^{-1}\pi_q(\beta) \in Z_q$. Which means we have well defined homomorphism

$\Pi_q: \Lambda \rightarrow \mathbb H (\mathbb{F}_q)^\times / Z_p$.

Let $T_{p,q} = \Pi_q(S_p)$, if $q > 2\sqrt{q}$ we have $\Pi_q$ is injective on $S_p$ and hence \$latex $| T_{p,q} | = p+1$.

Now we are ready to define our expanding family:

$X^{p,q} = \mathcal{G}( \Pi_q(\Lambda), T_{p,q})$.

Since $S_p$ generates $\Lambda$, $T_{p,q}$ generates $\Pi_q(\Lambda)$. Hence $X^{p,q}$ is $(p+1)$-regular and connected.

Theorem 1: $g(X^{p,q}) \geq 2 \log_p(q)$

Let $(e, v_1, \cdots, v_k=e)$ be a cycle in $X^{p,q}$, there is $t_1, \cdots, t_k \in T_{p,q}$ such that $v_i = t_1 t_2 \cdots t_k$ for $1 \leq i \leq k$.

Let $\gamma_i = \Pi_q^{-1}(t_i), \ \gamma_i \in S_p$, $\alpha = a_0 + a_1 i+a_2 j +a_3 k = \gamma_1 \cdots \gamma_k \in \Lambda$. Note that from the above arguement we know $\alpha$ is a reduced word, hence $\alpha \neq e_{\Lambda}$. In particular, this implies $a_1, a_2, a_3$ cannot all be $0$.

Also, since $\alpha$ is reduced, $\displaystyle N(\alpha) = \Pi_{i=1}^k N(\gamma_i) = p^k$.

By Lemma, since $\Pi_q(\alpha) = 1$, $\alpha \in \mbox{ker}(\Pi_q)$ hence $q$ divide $a_1, a_2, a_3$, we conclude

$N(\alpha) = a_0^2 + a_1^2 +a_2^2 +a_3^2 \geq q^2$

We deduce $p^k \geq q^2$ hence $k \geq \log_p(q)$ for all cycle. i.e. $g(X^{p,q}) \geq \log_p(q)$.

Theorem 2: If $q$ does not divide $p$ and $p$ is not a square mod $q$ (i.e. $(\frac{p}{q}) = -1$), then $g(X^{p,q}) \geq 4 \log_p(q) - \log_p(4)$.

For any cycle of length $k$ as above, we have $N(\alpha) = p^k \equiv a_0^2 (\mbox{mod} \ q)$, i.e. $(\frac{p^k}{q}) = 1$.

Since $(\frac{p^k}{q}) = ((\frac{p}{q})^k$, we have $(-1)^k = 1, \ k$ is even. Let $k = 2l$.

Note that $p^k \equiv a_0^2 (\mbox{mod} \ q^2)$, we also have

$p^{2l} \equiv a_0^2 (\mbox{mod} \ q^2)$

Hence $p^l \equiv a_0 (\mbox{mod} \ q^2)$.

Since $a_0^2 \leq p^{2l}$, $|a_0| \leq p^l$.

If $2l < 4 \log_p(q) - \log_p(4)$ we will have $p^l < q^2/2$. Then $|p^l\pm a_0| \leq 2|p^l| < q^2$.

But we know that $p^l \equiv a_0 (\mbox{mod} \ q^2)$, one of $p^l\pm a_0$ must be divisible by $q^2$, hence $0$.

Conclude $p^l = \pm a_0$, $N(\alpha) = p^k = a_0^2$, hence $a_1=a_2=a_3=0$. Contradiction.

# Isoperimetric inequality on groups

Back in high school, we have all learned and loved the isoperimetric inequality: “If one wants to enclose a region of area $\pi$ on the plane, one needs a rope of length at least $2 \pi$.” or equivalently, given $U \subseteq \mathbb{R}^2$ bounded open, we always have

$\ell(\partial U) \geq 2 \sqrt{\pi} \sqrt{\mbox{Area}(U)}$

Of course this generalizes to $\mathbb{R}^n$:

Theorem: Given any open set $U \subseteq \mathbb{R}^n$, we have

$\mbox{vol}_{n-1}\partial (U) \geq n\cdot \omega_n^{1/n} \mbox{vol}_n(U)^{\frac{n-1}{n}}$

Here $\omega_n$ is the volume of the unit n-ball. Note that the inequality is sharp for balls in $\mathbb{R}^n$.

One nature question to ask should be: for which other kind of spaces do we have such an inequality. i.e. when can we lower bound the volume of the boundary by in terms of the volume of the open set? If such inequality exists, how does the lower bound depend on the volume of the set?

I recently learned to produce such bounds on groups:
Let $G$ be a discrete group, fix a set of generators and let $C_G$ be its Cayley graph, equipped with the word metric $d$.

Let $N(R) = |B_R(e)|$ be the cardinality of the ball of radius $R$ around the identity.

For any set $U \subseteq G$, we define $\partial U = \{g \in G \ | \ d(g, U) = 1 \}$ i.e. points that’s not in $U$ but there is an edge in the Cayley graph connecting it to some point in $U$.

Theorem:For any group with the property that $N(R) \geq c_n R^n$, then for any set $U \subseteq G$ with $|U| \leq \frac{1}{2}|G|$,

$|\partial U| \geq c_n |U|^{\frac{n-1}{n}}$.

i.e. If the volume of balls grow (w.r.t. radius) as fast as what we have in $\mathbb{R}^n$, then we can lower bound the size of boundary any open set in terms of its volume just like what we have in $\mathbb{R}^n$.

Proof: We make use of the fact that right multiplications by elements of the group are isometries on Cayley graph.

Let $R = (\frac{2}{c_n}|U|)^\frac{1}{n}$, so we have $|B_R(e)| \geq 2|U|$.

For each element $g \in B_R(e)$, we look at how many elements of $U$ went outside of $U$ i.e. $| Ug \backslash U|$. (Here the idea being the size of the boundary can be lower bounded in terms of the length of the translation vector and the volume shifted outside the set. Hence we are interested in finding an element that’s not too far away from the identity but shifts a large volume of $U$ outside of $U$.)

The average value of $|Ug \backslash U|$ as $g$ varies in the ball $B_R(e)$ is:

$\displaystyle \frac{1}{|B_R(e)|} \sum_{g\in B_R(e)} |Ug \backslash U|$

The sum part is counting the elements of $U$ that’s translated outside $U$ by $g$ then sum over all $g \in B_R(e)$, this is same as first fixing any $u \in U$, count how many $g$ sends $u$ outside $U$, and sum over all $u \in U$ ( In fact this is just Fubini, where we exchange the order of two summations ). “how mant $g \in B_R(e)$ sends $u$ outside of $U$” is merely $|uB_R(e) \backslash U|$.

Hence the average is

$\displaystyle \frac{1}{|B_R(e)|} \sum_{u\in U}|uB_R(e) \backslash U|$.

But we know $|B_R(e)| \geq 2 \cdot |U|$, hence $|uB_R(e) \backslash U|$ is at least $\frac{1}{2}|B_R(e)|$.

Hence

$\displaystyle \frac{1}{|B_R(e)|} \sum_{u\in U}|uB_R(e) \backslash U|$

$\geq \frac{1}{|B_R(e)|} \cdot |U| \cdot \frac{1}{2} |B_R(e)| = \frac{1}{2} |U|$

Now we can pick some $g \in B_R(e)$ where $|Ug \backslash U| \geq \frac{1}{2}|U|$ (at least as large as average).

Since $g$ has norm at most $R$, we can find a word $g_1 g_2 \cdots g_k = g, \ k \leq R$.

For any $ug \in (Ug \backslash U)$, since $u \in U$ and $ug \notin U$, there must be some $1\leq i\leq k$ s.t. $u(g_1\cdots g_{i-1}) \in U$ and $u(g_1\cdots g_i) \notin U$.

Hence $u g_1 \cdots g_i \in \partial U$, $ug \in \partial U \cdot g_{i+1} \cdots g_k$.

We deduce that $\displaystyle Ug \backslash U \subseteq \partial U \cup \partial U g_k \cup \cdots \cup \partial U g_2 \cdots g_k$ i.e. a union of $k$ copies of $\partial U$.

Hence $R |\partial U| \geq k |\partial U| \geq |Ug \backslash U| \geq \frac{1}{2}|U|$

$|\partial U| \geq \frac{|U|}{2R}$

Since $|B_R(e)| \geq c_n R^n$, we have $R \leq c_n |B_R(e)|^{\frac{1}{n}}$. $R = (\frac{2}{c_n}|U|)^\frac{1}{n} = c_n |U|^\frac{1}{n}$, hence we have

$|\partial U| \geq c_n |U|^\frac{n-1}{n}$

Establishes the claim.

Remark: The same argument carries through as long as we have a lower bound on the volume of balls, not necessarily polynomial. For example, on expander graphs, the volume of balls grow exponentially: $B_R(e) \geq c \cdot e^R$, then trancing through the argument we get bound

$|\partial U| \geq c \cdot \frac{|U|}{\log{|U|}}$

Which is a very strong isoperimetric inequality. However in fact the sharp bound for expanders is $|\partial U| \geq c \cdot |U|$. But to get that one needs to use more information of the expander than merely the volume of balls.

On the same vein, we can also prove a version on Lie groups:

Theorem:Let $G$ be a Lie group, $g$ be a left invariant metric on $G$. If $\mbox{vol}_n(B_R) \geq c_n R^n$ then for any open set $U$ with no more than half of the volume of $G$,

$\mbox{vol}_{n-1}(\partial U) \geq c_n \mbox{vol}_n(U)^\frac{n-1}{n}$.

Note that to satisfy the condition $\mbox{vol}_n(B_R) \geq c_n R^n$, our Lie group must be at least n-dimensional since if not the condition would fail for small balls. $n$ might be strictly larger than the dimension of the manifold depending on how ‘neigatively curved’ the manifold is in large scale.

Sketch of proof: As above, take a ball twice the size of the set $U$ around the identity, say it’s $B_R(e)$. Now we consider all left translates of the set $U$ by element in $B_R(e)$. In average an element shifts at least half of $U$ outside of $U$. Pick element $g$ where $\mbox{vol}(gU \backslash U)$ is above average.

Let $\gamma: [0, ||g||]$ be a unit speed geodesic connecting $e$ to $g$. Consider the union of left translates of $\partial U$ by elements in $\gamma([0, ||g||])$, this must contain all of $gU \backslash U$ since for any $gu \notin U$ the segment $\gamma([0,||g||]) \cdot u$ must cross the boundary of $U$, i.e. there is $c \in [0,||g||]$ where $\gamma(c) u \in \partial U$, hence

$g\cdot u = \gamma(||g||) \cdot u = \gamma(||g||-c)\gamma(c) u \in \gamma(||g||-c) \cdot \partial U$

But since the geodesic has derivative $1$, the n-dimensional volume of the union of those translates is at most $\mbox{vol}_{n-1}(\partial U) \cdot ||g||$.

We have $\mbox{vol}_{n-1}(\partial U) \cdot R \geq \mbox{vol}_n(gU \backslash U) \geq \mbox{vol}_n(U)/2$

Now since we have growth condition

$2\mbox{vol}_n(U) = \mbox{vol}_n(B_R) \geq c_n R^n$

i.e. $R \leq c_n \mbox{vol}_n(U)^\frac{1}{n}$.

Conbine the two inequalities we obtain

$\mbox{vol}_{n-1}(\partial U) \geq c_n \mbox{vol}_n(U)^\frac{n-1}{n}$.

Establishes the theorem.

Remark: In general, this argument produces a lower bound on the size of the boundary in terms of the volume of the set as long as:
1. There is a way to ‘continuously translate’ the set by elements in the space.
2. We have a lower bound on the growth of balls in terms of radius.

The key observation is that the translated set minus the original set is always contained in a ‘flattened cylinder’ of the boundary in direction of the translate, which then has volume controlled by the boundary size and the length of the translating element. Because of this, the constant is almost never strict as the difference (translate subtract original) can never be the whole cylinder (in case of a ball, this difference is a bit more than half of the cylinder).